Question

Express the hcf of 48 and 18 as a linear combination

Answer 1

A=bq+r, where o ≤ r < b

48=18x2+12

18=12x1+6

12=6x2+0

∴ HCF (18,48) = 6

now 6= 18-12x1

6= 18-(48-18x2)

6= 18-48x1+18x2

6= 18x3-48x1

6= 18x3+48x(-1)

i.e. 6= 18x +48y

∴ x=3 , y=-1

6= 18×3 +48×(-1)

=18×3 +48×(-1) + 18×48-18×48

=18(3+48)+48(-1-18)

=18×51+48×(-19)

6=18x+48y

∴ x = 51, y = -19

Hence, x and y are not unique.